Posted by Shreya on Tuesday, October 2, 2012 at 1:34am.
does the following function have a hole or a vertical asymptote or both? how do you know? find the y-value at that point.
I factored this out to
y = (x+3)(x+4)/(x+5)(x+3)
The x+3's cancelled out. I dont get the whole verticle and horizontal asymptote stuff, how do u know w/o using technology whether there is an asymptote and if there is a hole?
- Pre-Calculus - Reiny, Tuesday, October 2, 2012 at 10:42am
If a pair of factors cancel, then you would get a hole
in your case, there is a hole when x = -3
If you have a factor in the denominator which does not cancel, it will cause a vertical asymptote
in your case there will be a vertical asymptote at x = -5
BTW, you should use brackets when typing a function like yours
= (x+3)(x+4)/( (x+5)(x+3) )
- Pre-Calculus - Steve, Tuesday, October 2, 2012 at 10:44am
there is a vertical asymptote if the denominator is zero and the numerator is not zero
there is a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator
There is a slant asymptote if the degree of the numerator is 1 more than the degree of the denominator.
I can't believe these facts were not covered in your book.
y = (x+4)/(x+5) as long as x ≠ -3
If x = -3, y = 0/0 which is not defined. There is a "hole" at (-3,1/2)
So, you should be able to figure the asymptotes present.
Visit wolframalpha.com and type in
you can play around with lots of stuff there
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